# Limit of rational function as $x$ approaches $2$.

$$\lim_{x \to 2} \frac{3x^2-13}{x-2}$$ How can this be done?Putting $$x=2$$ also doesn't give us $$\frac{0}{0}$$ form for which we could apply L'hopitals rule.Any ideas?

• isn't this just infinity? Maybe you wanted to write -12 then it becomes $3(x^2-4)$ which can be written as $3(x-2)(x+2)$ Jul 15 at 19:29

$$\lim_{x\rightarrow2^+}\frac{3x^2-13}{x-2}=\frac{-1}{0^+}=-\infty$$ and $$\lim_{x\rightarrow2^-}\frac{3x^2-13}{x-2}=\frac{-1}{0^-}=\infty$$. So this limit doesn't exist.
• I downvoted this answer because $\dfrac{-1}{0}$ is not defined. If $f(x)\to-1$ and $g(x)\to0$ as $x \to a^+$, and there is a $\delta>0$ such that if $0<x-a<\delta$ then $g(x)$ is positive, then $\dfrac{f(x)}{g(x)}\to\infty$ as $x\to a^+$; simply saying that $\dfrac{-1}{0}=-\infty$ is too sloppy in my opinion.
• @Joe Yes. You are right. Rigorously, we should use $\epsilon$-$\delta$ language. However, this is an intuitive result. When f(a) is not zero and g(a) is zero, you can always use this technic to see the limit quickly. Jul 17 at 1:29